Water flows from reservoir \"A\" at elevation 1500.00 ft. to reservoir \"E\" at

Question

Water flows from reservoir "A" at elevation 1500.00 ft. to reservoir "E" at elevation 500.00 ft. driving the motor shown. Head losses are as indicated on the figure. Do not consider any other losses. All piping is 12 inches in inside diameter. The pressure at "B" which has an elevation of 450.00 ft. is 400 psi. Find the discharge Q and the head removed by the motor). (Hint: use the Gen En eqn twice, once between A and E and then between A and B). Find: The discharge in cfs. The head removed by the motor in feet.

Explanation / Answer

Pressure at B = 400 psi = 400*122 lb/ft2 = 400*122/62.4 ft of water = 923 ft of water.

Applying steady flow energy equation between A and E,

(Pa/g + Va2/2g + za) = (Pe/g + Ve2/2g + ze) + (38V2/2g + 40V2/2g) + Hm

But Pa = Pe and Va = Ve 0.

Thus, Head taken away by motor, Hm = -(38V2/2g + 40V2/2g) + (1500-500) = 1000 - 78V2/2g

By applying steady flow energy equation between points A and B,

(Pa/g + Va2/2g + za) = (Pb/g + Vb2/2g + zb) + 40V2/2g

Putting Pa = 0 (gage) and Va 0, Pb/g = 923 ft, and Vb = V

So, 41V2/2g = 1500 - 923 - 450 = 127

So, 41V2/2g = 127

So, V = 14.1 ft/s

(a) So flowrate Q = /4*(12/12)2*14.1 = 11.09 ft3/s

(b) Hm = 1000-78*14.12/(2*32.2) = 759.2 ft

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